(i) The work done against friction is calculated using the formula:

\(\text{Work done against friction} = F \times d\)

where \(F = 40 \text{ N}\) and \(d = 36 \text{ m}\).

\(\text{Work done against friction} = 40 \times 36 = 1440 \text{ J}\)

(ii) The change in gravitational potential energy (PE) is given by:

\(\Delta PE = mgh\)

where \(m = 25 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 36 \sin 20^{\circ}\).

\(\Delta PE = 25 \times 9.8 \times 36 \times \sin 20^{\circ} \approx 3080 \text{ J}\)

(iii) The work done by the pulling force is the sum of the work done against friction and the change in gravitational potential energy:

\(\text{Work done by pulling force} = 1440 + 3080 = 4520 \text{ J}\)

(i) Gain in kinetic energy (KE) is given by:

\(\frac{1}{2} \times 1250 \times (8^2 - 5^2)\)

Loss in potential energy (PE) is given by:

\(1250g \times 400 \sin 4^\circ\)

Using work done (WD) by driving force (DF):

\(400(DF) = \frac{1}{2} \times 1250 \times (8^2 - 5^2) - 1250g \times 400 \sin 4^\circ + 2000 \times 400\)

Solving gives the driving force \(DF = 1189 \text{ N or } 1190 \text{ N}\).

(ii) Using Newton’s second law to find acceleration:

\(1189 \times 2 - 2000 = 1250a\)

Solving gives acceleration \(a = 0.302 \text{ m s}^{-2}\).

(iii) Using the formula for power:

\(v_c^2 = 64 + 2 \times 0.302 \times 750\)

\(P / 22.75 - 2000 = 1250 \times 0.302\)

Solving gives power \(P = 54.1 \text{ kW or } 54100 \text{ W}\).

(i) The work done by the car's engine is calculated using the formula for power:

\(P = \frac{\text{Work Done}}{\Delta t}\)

Given \(P = 14,000 \text{ W}\) and \(\Delta t = 25 \text{ s}\),

\(\text{Work Done} = 14,000 \times 25 = 350,000 \text{ J}\).

(ii) To find the gain in kinetic energy, first calculate the velocities at A and B using the formula:

\(\frac{P}{v} - 235 = 1600 \times a\)

For point A: \(\frac{14,000}{v_A} - 235 = 1600 \times 0.5\)

\(v_A = \frac{2800}{207} \approx 13.53 \text{ m/s}\)

For point B: \(\frac{14,000}{v_B} - 235 = 1600 \times 0.25\)

\(v_B = \frac{2800}{127} \approx 22.05 \text{ m/s}\)

The gain in kinetic energy is:

\(\text{KE gain} = \frac{1}{2} m (v_B^2 - v_A^2)\)

\(= \frac{1}{2} \times 1600 \times (22.05^2 - 13.53^2)\)

\(= 242,500 \text{ J}\) or \(242.5 \text{ kJ}\).

(iii) The distance \(AB\) is found using the work-energy principle:

\(350,000 = 242,500 + 235 \times AB\)

\(235 \times AB = 350,000 - 242,500\)

\(AB = \frac{107,500}{235} \approx 457 \text{ m}\).

First, calculate the frictional force along BC:

\(F = \mu R = 0.4 \times 2 \cos 45^{\circ} = 0.4 \sqrt{2}\)

Next, use energy conservation from A to C:

\(\text{KE gain} = \frac{1}{2} \times 0.2 \times V_C^2\)

\(\text{PE loss} = 0.2 \times g \times (2.5 + 2 \sqrt{2})\)

Set up the energy equation:

\(0.1 V_C^2 = 0.2 \times g \times (2.5 + 2 \sqrt{2}) - 0.4 \sqrt{2} \times 4\)

Solve for \(V_C\):

\(V_C = 9.16 \text{ m/s}\)

(i) The increase in kinetic energy \(\Delta KE\) is given by:

\(\Delta KE = \frac{1}{2} \times 8 \times 4.5^2 = 81 \text{ J}\)

The decrease in potential energy \(\Delta PE\) is given by:

\(\Delta PE = 8g \times 12 \times \left(\frac{1}{8}\right) = 120 \text{ J}\)

(ii) Using the energy principle, the work done by the resisting force \(R\) is:

\(81 = 120 - 12R\)

Solving for \(R\):

\(12R = 120 - 81\)

\(12R = 39\)

\(R = \frac{39}{12} = 3.25 \text{ N}\)